Problem 47 - Thomas Calculus

Question

Exercises \(45-48\) give equations for parabolas and tell how many units up or down and to the right or left each parabola is to be shifted. Find an equation for the new parabola, and find the new vertex, focus, and directrix. $$ x^{2}=8 y, \quad \text { right } 1, \text { down } 7 $$

Step-by-Step Solution

Answer

New equation: \((x-1)^2 = 8(y+7)\), Vertex: \((1, -7)\), Focus: \((1, -5)\), Directrix: \(y = -9\).

Step 1: Identify the Original Equation and Converting Intentions

The original equation given is \( x^2 = 8y \), which represents a parabola that opens upwards. We need to shift this parabola 1 unit to the right and 7 units down.

Step 2: Apply Horizontal and Vertical Shifts to the Equation

To shift the parabola horizontally by 1 unit to the right, replace \( x \) with \( (x-1) \). To shift it 7 units downwards, replace \( y \) with \( (y+7) \). Substituting these in, the new equation becomes: \( (x-1)^2 = 8(y+7) \).

Step 3: Find the New Vertex of the Parabola

The original vertex of \( x^2 = 8y \) is at the origin \((0, 0)\). After shifting, the new vertex is at \((1, -7)\) because we moved the parabola 1 unit right and 7 units down.

Step 4: Find the Focus of the New Parabola

The original parabola has a focus at \((0, 2)\) because \(4p = 8\) gives us \(p = 2\) (the distance from the vertex to the focus). After shifting, the new focus is at \((1, -5)\) because we move the focus the same amount as the vertex.

Step 5: Determine the New Directrix

The original directrix is at \(y = -2\), which is \(p\) (2) units below the vertex. After shifting down, the new directrix becomes \(y = -9\), which is 7 units below the shifted vertex.

Key Concepts

Vertex Shift

A vertex shift involves moving the vertex of a parabola to a new location in the coordinate plane. Here, our original parabola equation is \( x^2 = 8y \), which opens upwards with its vertex at the origin \((0, 0)\). To shift the vertex:

• Right by 1 unit: Substitute \( x \) with \( (x-1) \)
• Down by 7 units: Substitute \( y \) with \( (y+7) \)

This transformation results in \((x-1)^2 = 8(y+7)\) for the new equation. Thus, the new vertex position is \((1, -7)\), precisely reflecting the described shifts.

Focus and Directrix Calculation

Understanding a parabola's focus and directrix involves knowing how they relate geometrically to the vertex. Originally with \( x^2 = 8y \), the focus is calculated based on the formula \( 4p = 8 \) which simplifies to \( p = 2 \). Hence, the focus of the original parabola is at \((0, 2)\). After executing the same transformations applied to the vertex, the new focus is at \((1, -5)\).
The directrix complements the focus. Situated \( p \) units in the opposite direction of the vertex, the original directrix is at \( y = -2 \). Applying our downward shift translates this to \( y = -9 \) in the transformed equation. This consistent adjustment ensures our parabola maintains its geometric properties relative to the newly calculated positions.

Coordinate Transformation

Coordinate transformations apply systematic changes to the variables \( x \) and \( y \) to reflect specific movements of the graph in the coordinate plane. This principle underlies all changes in vertex, focus, and directrix.
Let's break it down:

• For a rightward shift, the operation involves \( x \rightarrow x-1 \). This transformation moves every point on the parabola exactly one unit to the right.
• A downward shift involves \( y \rightarrow y+7 \). In this case, every point on the parabola moves seven units downward.

These transformations are straightforward but crucial in adjusting parabolic equations to reflect desired positional changes accurately. With consistent application, they offer a robust method to adapt a parabola's positioning on a graph.